In Δ ABC, the perpendiculars drawn from A, B and C meet the opposite sides at points D, E and F, respectively. AD, BE and CF intersect at point P. If ∠ EPD = 110° and the bisectors of ∠ A and ∠ B meet at point Q, then ∠ AQB = ?
Answer & explanation
Correct answer: option 4
As we know,
⇒ \(\angle\)C + \(\angle\)EPD = \({180}^\circ\)
⇒ \(\angle\)C = \({180}^\circ\) - \({110}^\circ\) = \({70}^\circ\)
Again, as we know
⇒ \(\angle\)AQB = \({90}^\circ\) + \(\angle\)C/2
⇒ \(\angle\)AQB = \({90}^\circ\) + 70/2 = \({90}^\circ\) + \({35}^\circ\) = \({125}^\circ\).